Chapter 7 Notes and Problem Set

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					                       Chapter 7 Notes and Problem Set (memorize items in bold letters)

A. Internal energy measured in joules (J) or calories. 4.184J = 1 cal The internal energy, E, of a system is
the sum of the kinetic and potential energies of all the “particles: in the system. The E of the system can be
changed by a slow of work (w), heat (q) or both.

      E = q + work or E = q – PV because w = – PV , where q, P, and V are heat, pressure and
volume respectively.

Heat is the energy that is transferred from the object’s internal energy, E.
               In a chemical process: E = Efinal – Einitial = Eproducts - Ereactants

                         E is + if the system (the chemical process) absorbs energy from the surroundings and
                         E is “-“ if the system releases energy to its surroundings

A.1. Heat of combustion is measured in a bomb calorimeter because the reactions consume and produce
gases. In bomb calorimeter: V = 0, therefore, E = qv where qv is q at constant volume. Heat is
proportional to T, where T is the temperature change. q = CT where: C = heat capacity in J/oC (is an
extensive property which is proportional to sample mass) and T = Tfinal – Tinitial.

Sample problems:

1a.1. 1.000 g olive oil is completely burned in pure oxygen in a bomb calorimeter. The temperature of the
water bath increases from 22.000oC to 22.241oC. How many dietary calories are in olive oil per gram? The
heat capacity is 9.032 kJ/oC. Answer: -0.521Cal/g; to solve the problem you need to use: q = CT and use
4.184kJ = 1 kcal = 1 Cal; Note: if q of the water bath is “+” q since the water absorbed q and therefore, olive
oil is” –“ since qH2O = -qsubstance)

1a.2. olive oil is almost pure glyceryl trioleate, C57H104O6. The equation for the combustion of glyceryl
trioleate is: C57H104O6 + 80 O2  57CO2 + 52 H2O. What is the change in the internal energy, E, in kJ for
the combustion of 1 mol glyceryl trioleate? Answer: -1.93kJ/mol glyceryl trioleate; use E = qv where qv =
CT from 1.a.1. i.e. -2.18kJ/1g x 885.4 g C57H104O6/mol = -1.93kJ/mol.

2b. (Practice exercise 6) A 1.50 g sample of C is burned in a bomb calorimeter that has a heat capacity of
8.93 kJ/ oC. The temperature of the water jacket rises from 20.00oC to 25.51oC. What is the E for the
combustion of 1 mol C? Answer: -394kJ/mol C; E= qv = CT.
A.2. For reactions other than combustion in a bomb calorimeter:

          Heat = q = mST where m = mass in grams, S = specific heat in J/goC (intensive property), and T =
Tfinal – Tinitial. = A2 equation; Swater = 4.184 J/goC.

Sample problems:

2a. How many joules are needed to increase the temperature of 15.0 g Fe from 20.0 to 40.0oC? (Specific
heat of Fe = 0.4498 J/goC). Answer: 135J; use the A.2 equation.

2b. The addition of 250J to 30.0g Cu initially at 22oC will change its temperature to what final value? Answer:
44oC; use the A2 equation.

2c. In a two system reaction like hot metal submerged in a water bath in a calorimeter th q absorbed by
water is a positive value and q lost by the metal is a negative value so q H2O = -qmetal. Tfinal is the temperature
at which both the metal and water reach thermal equilibrium, i.e. they are at the same temperature.

A sample of Cu was heated to 120 oC and then thrust into 200.g water at 25.00 oC. The temperature of the
mixture became 26.50 oC.

2c1. How much heat in J was absorbed by the water? Answer: 1.26 x 103 J; use A2 equation and 4.184J/goC
for S of water.

2c2. The copper sample lost how many J? Answer: 1.26 x 103 J; qCu = -1.26 x 103 J

2c3. What was the mass in grams of the Cu sample (SCu = 0.387 J/g oC)? Answer: 34.7g Cu; use A2
equation and substitute qCu = -1.26 x 103 J, T = 120-26.5 oC and SCu = 0.387 J/g oC and solve for m.
B. Constant pressure calorimetry: Heat of reaction in solution-Chemical bonds

1. breaking a chemical bond requires energy and making a chemical bond releases energy. The net gain or
loss of energy is often in the form of heat. (in exothermic reactions heat is a product, i.e. the reaction
releases heat, H = Hproducts - Hreactants is negative; in endothermic reactions heat is a reactant and H is

2. The amount of heat absorbed or released by a chemical reaction is usually measured in open containers at
constant pressure (P). Under these conditions E = qp + w = q – PV because w = – PV. PV is zero for
liquids and solids.

Therefore, qp = E + PV = H. I.e. qp = H = Hproducts - Hreactants = enthalpy= heat change at constant P.
The value of H is “+” for an endothermic reaction and H is “-“ for an exothermic reaction.

3. From calculations of q can find H of a reaction:

        e.g. Suppose we mix 50.0 mL of 1.0 M HCl at 25.0oC with 50.0 mL of 1.0 M NaOH also at 25oC in a
calorimeter. After the reactants are mixed by stirring, the temperature is observed to increase to 31.9 oC. In
this neutralization reaction: HCl + NaOH  NaCl + H20; net ionic equation: H+ + OH-  H20. Assume
water has a density of 1.000 g/mL. What is the H/mol HCl?

The qH20 can be calculated so qreaction = – qH20.

       qH20 = m x T x S = where m = mass of total system in g; T = Tfinal – Tinitial; SH20 = 4.184 J/g oC
            = 100.0g x 6.9 oC x 4.184 J/g oC = 2.9 x 103 J.

qreaction = -2.9 x 103 J for 0.05L x 1.0 M HCl (0.05 mol). Thus, H/mol HCl = -5.8 J/mol.

4. The standard enthalpy, Ho, under standard conditions of 1 bar (IUPAC) or 1 atmosphere (earlier
definition) pressure for all gases, 1 M concentration for aqueous solutions and is often at 25oC or 298K) for
compounds. The standard state of an element is the form in which the element exists under conditions of 1
atmosphere and 25oC (for example the standard state for oxygen is O2 (g); for sodium is Na (s); for mercury
is Hg (l)).

5. The enthalpy changes for reactions can be calculated by algebraic summation based on Hess’s law:

Rules for Hess’s law

1. When an equation is reversed the sign of the enthalpy change must also be reversed.

2. Formulas canceled from both sides of an equation must be for substances in identical physical states.

3. If all the coefficients of an equation are multiplied or divided by the same factor, the value of the enthalpy
change must likewise be multiplied or divided by that factor.
Sample problems:

5a. Practice exercise 10: Ethanol, C2H5OH is made industrially by the reaction of water with ethylene, C2H2.
Calculate the Ho for the reaction: C2H2 (g) + H2O (l)  C2H5OH (l) given the following thermochemical
               C2H2 (g) + 3O2 (g)  2CO2 (g) + 2H2O (l) Ho = -1411.1 kJ
               C2H5OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (l) Ho = -1367.1 kJ

Answer: Ho = -44.0 kJ

5b. Diborane (B2H6) is highly reactive boron hydride that was once considered as a possible rocket fuel.
Calculate H for the synthesis of diborane from its elements, according to the equation: 2B + 3H2  B2H6
using the following data:
(a) 2B (s) + 3/2 O2(g)  B2O3 (s)                    -1273 kJ
(b) B2H6 (g) + 3O2 (g)  B2O3 (s) + 3H2O             -2035 kJ
(c) H2 (g) + ½ O2 (g)  H2O (l)                      -286 kJ
(d) H2O (l)             H2O (g)                     44 kJ

Answer: H = +36kJ

6. Ho for the reaction can also be calculated from the standard enthalpy of formation (Hfo) using Hess’s
law. Hfo is where the elements in the reaction are in their standard states (see #4) and 1 mol of product is
formed. Hfo is reported per mol of product.

       Therefore: Horeaction = np(products) – nr(reactants) where  = sum of the terms; np and nr
       represent the moles of each product or reactant, respectively.

       NoteHo for an element in its standard state = 0.
6a. Practice exercise 13a: Calculate the Ho for the following reaction using the standard enthalpies in table
7.2 of your text: 2NO (g) + O2 (g)  2NO2 (g)

Answer: -113.1 kJ see solution’s guide

6b. Use the standard enthalpies in table 7.2 of your text to calculate the standard enthalpy change for the
overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water. This it the first
step in the manufacture of nitric acid.

       4NH3 (g) + 7O2 (g)  4NO2 (g) + 6H2O (l)

Answer: Horeaction = -1396 kJ

Solution: You will need the following reactions:
½ N2 (g) + 3/2 H2 (g)  NH3 (g)               -46 kJ/mol
½ N2 (g) + O2 (g)  NO2                       34 kJ/mol